5. Which of the following is/are assumptions in using F-test? А. There is homogeneity among its variance. В. The data are normally distributed. С. The groups are independent of each other. All of the above.
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- A behavioral researcher measures the stress response of mice exposed to scary animal pictures (cat, dog, lion, Cheshire cat) with a GABA sensor implanted in the brain. He uses randomly assigned 40 mice, 10 in each group. The researcher records the amount of stress after 30 seconds assuming that there is no difference in the response in whatever way the mouse is scared. The data is normally distributed. Means and variances are provided in the table below. Is there a difference in the response to scary animals. Group 1 - Cat n₁=10 m₁ = 10 s² 1= 1.8 Group 2 - Dog n2=10 m₂= 5.0 s²2= 3.4 Why is the scare effect so low in group 4? Group 3 - Lion n3=10 m3= 3.4 s²1= 2.5 Group 4 - Chesire Cat n4=10 m4= 2.4 s² 1= 1.6Find the standardized test statistic, z to test the claim that p, = p2. The sample statistics listed below are from independent samples. n1 = 50, x1 = 35, and n2 = 60, x2 = 40 ... О А. 1.328 В. 2.361 С. О.374 O D. 0.9820. A high school principal claims that the variance in the GPA of students in the graduating class is Tess than 0.39. You test this claim by sampling 31 seniors and find their GPA have a variance or 0.21. Ho:o? Ha:o2 < 0.39 (Claim) = 0.39 Is there enough evidence to reject Ho and support the principal's claim at a 0.10 level of significance? ohs A. No, the test statistic x? B. No, the test statistic x2 = 16.154 is not in the rejection region with critical value 40.256 C. Yes, the test statistic x? D. No, the test statistic x2 = 21.434 is not in the rejection region with critical value 20.599 E. Yes, the test statistic x? = 16.154 is in the rejection region with critical value 20.599 = 20.599 is not in the rejection region with critical value 21.434 %3D 20.599 is in the rejection region with critical value 16.154 %D
- Suppose that the blood pressure for a randomly selected patient has an average of 22 and variance of 441. If a sample of 220 patients is selected, then what is P(-1.2 < X< 1.4) ?3.56 An oil exploration firm is formed with enough capital to finance ten explorations. The probabil- ity of a particular exploration being successful is .1. Assume the explorations are independent. Find the mean and variance of the number of successful explorations.A company produces bags of refined sugar. We are interested in testing the hypothesis that the average weight of the bags is 20 kg. The weights of the contents of these bags are normally distributed with a population variance of 1.96 square kg. The contents of a random sample of 49 bags had a mean weight of 19.7 kg. Then the value of the test statistic is اختر أحد الخيارات a. -1.50 b. 1.50 O C. 1.07 O d. -1.07 O
- Give a 99.8% confidence interval, for μ₁ −μ₂ given the following information. n₁ = 35, x₁ = 1 2.41, s₁ = 0.39 81 n₂ = 20, x₂ = 2.62, s₂ = 0.83 2 -0.21 ✓0.69 X Use Technology Rounded to 2 decimal places.Suppose that 90% of Narnians are children, that Narnian children have normallydistributed BMIs with a mean of 25 kg/m2and variance of 64 (kg/m2) 2, and that Narnianadults have normally distributed BMIs with a mean of 30 kg/m2and variance of 100(kg/m2)2. WHO currently categorizes BMIs greater than 30 as obese.a. What is the prevalence of obesity among Narnian children?b. What is the prevalence of obesity among Narnian adults?c. What is the overall prevalence of obesity among Narnians?d. Are age and obesity statistically independent in Narnia?e. What proportion of Narnians are obese children?f. What proportion of obese Narnians are children?3. We want to compare initial salaries of graduates with Marketing major and graduates with Finance Major. We compare the salaries of these two groups with the same GPA. Example, salary of marketing graduate with GPA 3.5 is compare with the salary of Finance graduate with the same GPA 3.5. Assume salaries are normally distributed for both groups with unequal variances. We want to study the mean of the difference of these two groups. This is a a) t-test with equal variances b) t-test with unequal variances c) t-test for paired data.
- A transect is an archaeological study area that is mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance . In a different section of Chaco Canyon, a random sample of 23 transects gave a sample variance for the number of sites per transect. Use an to test the claim that the variance in the new section is greater than 36.4. What is the level of significance? 90% 1% 5% 95% 10%7. A study was conducted to compare the amount of salt in potato chips. Random samples of recorded. Assume the populations are normally distributed with equal variances. The 7 A study was conducted to compare the amount of salt in potato chips. Random sampies o. three varieties were obtained and the amount of salt in each 1-Oz portion of potato chp recorded. Assume the populations are normally distributed with equal variances. results are given here: BBQ Cheese- Amount of salt (in mg of sodium) 338, 159, 240, 190 235, 251, 233, 255, 260 Variety flavored Olestra-based 164, 150, 149, 170 a) Identify the explanatory variable. b) Identify the response variable. c) State the appropriate hypotheses to test for a difference in means. d) The appropriate analysis to test for a difference in means would be: a. One-way ANOVA b. Two-way ANOVA c. Three-way ANOVA d. Two-Sample t-testA transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance ?2 = 42.3. In a different section of Chaco Canyon, a random sample of 19 transects gave a sample variance s2 = 46.9 for the number of sites per transect. Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3. Find a 95% confidence interval for the population variance. (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)- 19.96 What are the degrees of freedom? -18 (c) Find or estimate the P-value of the sample test statistic. P-value > 0.1000.050 < P-value < 0.100 0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005…